OCR A Level Chemistry A: Acids, 2.1.4 (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 2.1.4, Acids, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. All five learning outcomes, (a) to (e), are tested. Use H 1.0, C 12.0, N 14.0, O 16.0, Na 23.0, Ca 40.1.

Read the 2.1.4 study guide and the 2.1.4 revision notes first. Course hub: OCR A Level chemistry. Tracker: printable checklist. If these feel hard, try the free diagnostics.

Questions

1. (a) Give the formulae of ethanoic acid and of potassium hydroxide. [2] (b) State the difference between a base and an alkali, and give the formula of one base that is not an alkali. [2]

2. Hydrochloric acid and ethanoic acid solutions have the same concentration. Explain, in terms of dissociation, why hydrochloric acid is strong and ethanoic acid is weak. Then explain why 25.0 cm³ of each solution needs the same volume of one sodium hydroxide solution for neutralisation. [4]

3. Write full equations for these reactions. (a) Copper(II) oxide with sulfuric acid. [1] (b) Potassium carbonate with nitric acid. [2] (c) Aqueous ammonia with sulfuric acid, forming a salt only. [1] (d) The ionic equation common to every acid and alkali neutralisation. [1]

4. Calculate the minimum volume of 1.00 mol dm⁻³ hydrochloric acid needed to react completely with 1.50 g of calcium carbonate. Give your answer in cm³ to 3 significant figures. [4]

5. A technician makes 100.0 cm³ of a 0.0800 mol dm⁻³ standard solution of ethanedioic acid from the solid hydrate, H₂C₂O₄·2H₂O. (a) Calculate the mass of the hydrate required. [3] (b) Describe how the solution is made, from the weighed solid to the finished solution. [3]

6. A student titrates 25.0 cm³ samples of an alkali with an acid in a burette. For each fault, state the effect on the titre (too high, too low, or no effect) and explain it. (a) The pipette was rinsed with distilled water, not with the alkali, before use. [2] (b) An air bubble in the burette jet at the start left the jet during the titration. [2] (c) The conical flask was rinsed with distilled water before the alkali was added. [1]

7. 25.0 cm³ portions of sulfuric acid of unknown concentration were titrated with 0.150 mol dm⁻³ sodium hydroxide. The titres were 26.80 (rough), 24.60, 24.50 and 24.55 cm³. (a) State the mean titre you would use and why. [2] (b) Write the equation for the reaction. [1] (c) Calculate the concentration of the sulfuric acid in mol dm⁻³, to 3 significant figures. [3]

8. Acid X is a solid that releases two H⁺ ions per molecule, written H₂X. A sample of 0.590 g of X was dissolved in water to give 250.0 cm³ of solution. One 25.0 cm³ portion needed 25.00 cm³ of 0.0400 mol dm⁻³ potassium hydroxide for neutralisation. Calculate the relative molecular mass of X. [6]

9. An antacid tablet of mass 2.00 g contains calcium carbonate as its only base. The tablet was crushed and 50.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid, an excess, was added. The mixture needed 20.00 cm³ of 0.500 mol dm⁻³ sodium hydroxide to neutralise the acid left over. Calculate the percentage by mass of calcium carbonate in the tablet. [7]

10. A 25.0 cm³ sample of a cleaning solution containing ammonia was diluted to 250.0 cm³. A 25.0 cm³ portion of the diluted solution needed 21.30 cm³ of 0.0500 mol dm⁻³ sulfuric acid. Calculate the concentration of ammonia in the cleaning solution in g dm⁻³, to 3 significant figures. [6]

11. A student compares 25.0 cm³ of 0.0100 mol dm⁻³ hydrochloric acid with 25.0 cm³ of 0.150 mol dm⁻³ ethanoic acid. Each is titrated with 0.100 mol dm⁻³ sodium hydroxide. (a) Calculate the volume of sodium hydroxide needed for each acid. [3] (b) The student concludes that ethanoic acid is the stronger acid because it needs more alkali. Explain why this conclusion is wrong. [3] (c) Write the equation for ethanoic acid with sodium hydroxide. [1] (d) State how many times more ethanoic acid than hydrochloric acid is present in the two samples. [1]

Answers

1. (a) CH₃COOH [1]; KOH [1] (b) An alkali is a base that dissolves in water (a soluble base) [1]; a base such as CuO (or MgO or CaCO₃) is not an alkali because it is insoluble [1] Examiner insight: Define an alkali as a soluble base. Saying “an alkali is a base” without the solubility point does not separate the two terms.

2. Hydrochloric acid is almost completely dissociated into H⁺ and Cl⁻ in water [1]; ethanoic acid is only partially dissociated, so most of it stays as CH₃COOH molecules in equilibrium with a little H⁺ [1]. Equal volumes of equal concentrations hold the same amount of acid, and each has one H⁺ to give per molecule [1]. As NaOH removes H⁺, more ethanoic acid dissociates, so all of it is eventually neutralised and the volume of NaOH is the same [1] Examiner insight: Strong and weak are about extent of dissociation; “higher pH” or “less reactive” does not explain it in those terms.

3. (a) CuO + H₂SO₄ → CuSO₄ + H₂O [1] (b) K₂CO₃ + 2HNO₃ → 2KNO₃ + H₂O + CO₂ [1] for correct formulae and products; [1] for balancing (c) 2NH₃ + H₂SO₄ → (NH₄)₂SO₄ [1] (d) H⁺(aq) + OH⁻(aq) → H₂O(l) [1] Examiner insight: Check the charge balance in the salt formula before balancing; a wrong formula cannot be rescued by balancing.

4. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ [1] n(CaCO₃) = 1.50 / 100.1 = 0.01499 mol [1] n(HCl) = 2 × 0.01499 = 0.02997 mol [1] V = 0.02997 / 1.00 = 0.02997 dm³ = 30.0 cm³ [1] Examiner insight: Show the 1 : 2 ratio line; it is the step easiest to drop.

5. (a) M(hydrate) = 24.0 + 2.0 + 64.0 + 36.0 = 126.0 g mol⁻¹, including the two waters [1]; n = 0.1000 dm³ × 0.0800 = 0.00800 mol [1]; m = 0.00800 × 126.0 = 1.01 g [1] (b) Weigh the solid accurately, dissolve it in a little distilled water in a beaker, and transfer it to a 100.0 cm³ volumetric flask, washing the beaker and funnel into the flask with distilled water [1]. Fill until the lowest point of the meniscus sits on the graduation mark, adding the last drops with a dropping pipette [1]. Stopper and invert several times to mix [1] Examiner insight: Use the molar mass of the hydrate, not of the anhydrous acid. Include the rinsing of the beaker and the meniscus on the mark.

6. (a) Titre too low [1]; water dilutes the alkali in the pipette, so the aliquot contains fewer moles of alkali and needs less acid [1] (b) Titre too high [1]; the volume of the bubble is counted as delivered acid, but no acid was delivered by it [1] (c) No effect [1], because the moles of alkali added to the flask are unchanged by water already in it Examiner insight: State the direction of the error and then the reason. A direction on its own, or a reason with no direction, is incomplete.

7. (a) Mean of 24.60, 24.50 and 24.55 = 24.55 cm³ [1]; the rough titre is excluded because it is not accurate, and the other three agree closely [1] (b) H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O [1] (c) n(NaOH) = 0.02455 × 0.150 = 3.68 × 10⁻³ mol [1]; n(H₂SO₄) = 3.6825 × 10⁻³ / 2 = 1.84 × 10⁻³ mol [1]; c = 1.84125 × 10⁻³ / 0.0250 = 0.0737 mol dm⁻³ [1] Examiner insight: Keep extra figures until the last line and round once.

8. n(KOH) = 0.02500 × 0.0400 = 1.00 × 10⁻³ mol [1] H₂X + 2KOH → K₂X + 2H₂O, so the ratio is 1 : 2 [1] n(H₂X) in 25.0 cm³ = 5.00 × 10⁻⁴ mol [1] In 250.0 cm³: 5.00 × 10⁻⁴ × 10 = 5.00 × 10⁻³ mol [1] M = 0.590 / 5.00 × 10⁻³ [1] = 118 g mol⁻¹ [1] Examiner insight: For an unfamiliar acid, write the equation first, and show the scale-up from aliquot to flask as a separate step.

9. n(HCl added) = 0.0500 × 0.500 = 0.0250 mol [1] n(NaOH) = 0.02000 × 0.500 = 0.0100 mol, equal to the HCl left over (1 : 1) [1] n(HCl reacted with tablet) = 0.0250 − 0.0100 = 0.0150 mol [1] CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ [1] n(CaCO₃) = 0.0150 / 2 = 0.00750 mol [1] m = 0.00750 × 100.1 = 0.751 g [1] Percentage = 0.751 / 2.00 × 100 = 37.5% [1] Examiner insight: In a back titration, subtract the leftover from the total before using the ratio; treating the titre as the acid that reacted with the sample loses credit.

10. 2NH₃ + H₂SO₄ → (NH₄)₂SO₄ [1] n(H₂SO₄) = 0.02130 × 0.0500 = 1.065 × 10⁻³ mol [1] n(NH₃) = 2 × 1.065 × 10⁻³ = 2.13 × 10⁻³ mol [1] c(diluted) = 2.13 × 10⁻³ / 0.0250 = 0.0852 mol dm⁻³ [1] c(original) = 0.0852 × 10 = 0.852 mol dm⁻³ [1] Mass concentration = 0.852 × 17.0 = 14.5 g dm⁻³ [1] Examiner insight: Write the dilution factor as a ratio of volumes, 250.0 / 25.0, and say which solution each concentration refers to.

11. (a) n(HCl) = 0.0250 × 0.0100 = 2.50 × 10⁻⁴ mol, so V(NaOH) = 2.50 × 10⁻⁴ / 0.100 = 2.50 cm³ [1] [1]; n(CH₃COOH) = 0.0250 × 0.150 = 3.75 × 10⁻³ mol, so V(NaOH) = 37.5 cm³ [1] (b) The volume of alkali depends on the amount of acid present, which is concentration × volume, and not on strength [1]. Ethanoic acid is the weak acid: it is only partially dissociated [1]. The hydrochloric acid is strong, completely dissociated, and needs less alkali only because it is much more dilute [1] (c) CH₃COOH + NaOH → CH₃COONa + H₂O [1] (d) 3.75 × 10⁻³ / 2.50 × 10⁻⁴ = 15 times [1] Examiner insight: Link “needs more alkali” to moles of acid, then say strength is a separate property.

Where marks are usually lost

• Describing a weak acid as “low concentration” or “less acidic” rather than “partially dissociated”.
• Leaving out CO₂ in a carbonate equation, or giving water as a product of ammonia with an acid.
• Missing the 1 : 2 ratio for sulfuric acid with a monohydroxide, or for calcium carbonate with hydrochloric acid.
• Averaging the rough titre in with the accurate ones.
• Forgetting to scale from the 25.0 cm³ aliquot to the full volume of solution made up.
• Using the anhydrous molar mass for a hydrate.
• Subtracting the wrong way in a back titration, or using titre moles directly.

Next steps

• 2.1.4 revision notes
• 2.1.4 study guide
• Amount of substance (2.1.3) for the mole calculations behind titrations
• Acids, bases and buffers for the quantitative treatment of weak acids
• OCR A Level chemistry hub
• Printable OCR chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Frequently asked questions

Are these OCR A Level Chemistry 2.1.4 Acids questions from past papers? No. They are original practice questions with worked answers on section 2.1.4 of OCR A Level Chemistry A (H432).

How should you describe a weak acid for the mark? Say that it is partially dissociated in water. Describing it as “low concentration” or “less acidic” does not earn the mark.

Should the rough titre be included in the mean? No. The rough titre is a guide only, so average just the accurate titres that agree closely.

What is a back titration mistake to avoid? Subtracting the wrong way round, or using the moles from the titre directly as the moles of the substance being analysed.

Why do you need to scale up in some titration answers? If only an aliquot, such as 25.0 cm³, of a made-up solution was titrated, multiply up to the full volume of solution that was made.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 2.1.4 Acids, learning outcomes (a) to (e), with additional guidance PAG2 and HSW4. Published by OCR.